Question

As Fig. 39-8 suggests, the probability density for the region X>L in the finite potential well of Fig. 39-7 drops off exponentially according to${\mathit{\psi }}^{\mathbf{2}}\left(x\right)\mathbf{=}\mathit{C}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{k}\mathbf{x}}$ , where C is a constant. (a) Show that the wave function$\mathit{\psi }\left(x\right)$ that may be found from this equation is a solution of Schrödinger’s equation in its one-dimensional form. (b) Find an expression for k for this to be true.

Short answer

a) It is proved that the wave function$\psi \left(x\right)$ is a solution of Schrödinger’s equation in its one-dimensional form.

b) The expression of is$k=\frac{2\mathrm{\pi }}{h}\sqrt{2m\left(U-E\right)}$ .

Step-by-step solution

Describe Schrodinger's equation

Schrodinger's equation in the finite potential well at is given by,

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{\psi }}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{8}{\mathbf{\pi }}^{\mathbf{2}}\mathbf{m}}{{\mathbf{h}}^{\mathbf{2}}}\left(E-U\right)\mathbf{\psi }\mathbf{=}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Show that the wave function ψ(x) is a solution of Schrödinger’s equation in its one-dimensional form(a)

Consider the given wave function.

$$\begin{gathered} {\psi }^2\left(x\right)=C{e}^{-2kx} \\ \psi \left(x\right)=\sqrt{C}{e}^{-kx} \end{gathered}$$

Substitute$\psi \left(x\right)=\sqrt{C}{e}^{-kx}$ in equation (1).

$$\begin{gathered} \frac{d^2}{d{x}^2}\left(\sqrt{C}{e}^{-kx}\right)+\frac{8{\mathrm{\pi }}^2\mathrm{m}}{h^2}\left(E-U\right)\psi =0 \\ \sqrt{C}\frac{d}{dx}\left(-k{e}^{-kx}\right)+\frac{8{\mathrm{\pi }}^2\mathrm{m}}{h^2}\left(E-U\right)\psi =0 \\ \sqrt{C}{k}^2{e}^{-kx}+\frac{8{\mathrm{\pi }}^2\mathrm{m}}{h^2}\left(E-U\right)\psi =0 \\ {k}^2\psi +\frac{8{\mathrm{\pi }}^2\mathrm{m}}{h^2}\left(E-U\right)\psi =0 \end{gathered}$$

Therefore, it is proved that the wave function$\psi \left(x\right)$ is a solution of Schrödinger’s equation in its one-dimensional form.

Find an expression for k for this to be true(b)

Consider the following equation.

$k^2\psi +\frac{8{\mathrm{\pi }}^2\mathrm{m}}{h^2}\left(E-U\right)\psi =0$

Simplify the above equation for .

$$\begin{gathered} k^2\psi =\frac{8{\mathrm{\pi }}^2\mathrm{m}}{h^2}\left(U-E\right)\psi \\ {k}^2=\frac{8{\mathrm{\pi }}^2\mathrm{m}}{h^2}\left(U-E\right) \\ k=\frac{2\mathrm{\pi }}{h}\sqrt{2m\left(U-E\right)} \end{gathered}$$

Therefore, the expression of is $k=\frac{2\mathrm{\pi }}{h}\sqrt{2m\left(U-E\right)}$.